Tangent planes

These two space curves are the graphs of vector functions $${\bf r}_1(x)\ = \ \big\langle\, x, \,b, \,f(x, b)\big \rangle\,,$$ $${\bf r}_2(y)\ = \ \big\langle \,a,\, y, \,f(a, y)\big \rangle,$$ shown in orange on the surface. The vector derivatives $${\bf T}_1 \ = \ {\bf r}_1'(a)\ = \ {\bf i}+ f_x(a, b)\,{\bf k}\,,$$ $${\bf T}_2\ = \ {\bf r}_2 '(b)\ = \ {\bf j}+ f_y(a, b)\,{\bf k}\,,$$ are Tangent Vectors at $P$ to the graph of $f$, while the plane containing $P$ as well as ${\bf T}_1$and ${\bf T}_2$ is the Tangent Plane at $P$.     To calculate the equation of the tangent plane at $P(a,\,b,\,c)$ we need its normal: $${\bf n} \ = \ {\bf T}_1 \, \times \, {\bf T}_2 \ = \ \left[\begin{array} {ccc} \,{\bf i} \,&\, {\bf j}\,& {\bf k } \\ 1 & 0 & f_x(a,\,b) \\ 0 & 1 & f_y(a,\,b) \end{array}\right]$$ $$\qquad = \ \big \langle\, - f_x(a,\,b),\, - f_y(a,\,b),\,1\, \big \rangle\,.$$

    On the other hand, if $Q(x, \,y,\, z)$ is an arbitrary point in the tangent plane at $P(a,\,b,\,c)$, then $$\overrightarrow{PQ}\ = \ \big \langle\, x - a,\, y - b,\, z - f(a, b)\, \big \rangle$$ lies in the plane and so is perpendicular to ${\bf n}$. Thus in point-normal form the equation of the tangent plane is $$\big \langle\, x - a,\, y - b,\, z - f(a, b)\, \big \rangle \cdot {\bf n} \ = \ \big \langle\, x - a,\, y - b,\, z - f(a, b)\, \big \rangle \cdot \big \langle\, - f_x(a,\,b),\, - f_y(a,\,b),\,1\, \big \rangle$$ $$= \ -(x-a)f_x(a, b) - (y-b)f_y(a, b) + z - f(a, b) \ = \ 0\,. \qquad \qquad$$ After rearranging we get:

The equation of the Tangent Plane at $(a,\, b,\, f(a, \,b))$ is $$z\ = \ f(a, b) + (x-a)f_x(a, b) + (y-b)f_y(a, b)\,.$$